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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
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If G is the universal gravitational constant and $\rho$ is the uniform density of a spherical planet. Then the shortest possible period of rotation of the planet can be

\[(a)\;\sqrt {\frac{\pi G}{2 \rho}} \quad (b)\;\sqrt {\frac{3 \pi G}{\rho}} \quad (c)\;\sqrt {\frac{\pi}{6 G \rho}} \quad (d)\;\sqrt {\frac{3 \pi}{G\rho}} \]
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1 Answer

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For time period to be shortest $w_{max}$ the fastest possible rate of rotation of a planet for which gravitational force on the equator just barely provides the centripetal force needed for the rotation.
Let M be mass and R the radius of planet.
$mRw_{max}^2=\large\frac{GMm}{R^2}$
$w_{max}=\sqrt {\large\frac{GM}{R^3}}$
$\qquad= \sqrt {\large\frac{G\bigg(\Large\frac{4}{3} \pi R^3 \rho\bigg)}{R^3}}$
$\qquad=2 \sqrt {\large\frac{\pi G \rho}{3}}$
$\large\frac{2 \pi}{T_{ \Large min}}=2 \sqrt {\large\frac{\pi G \rho}{3}}$
$T_{ min} = \sqrt {\large\frac{3 \pi}{G \rho}}$
Hence d is the correct answer.

 

answered Aug 30, 2013 by meena.p
edited Feb 18, 2014 by meena.p
 

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